Question

Find the third degree polynomial function that has an output of 40 when x=1, and has zeros −19 and −i?

Answer 1

Answer:

$f(x) = (x + 19)(x^2 + 1)$

Step-by-step explanation:

Complex numbers:

The following relation is important for complex numbers:

$i^2 = -1$

Zeros of a function:

Given a polynomial f(x), this polynomial has roots $x_{1}, x_{2}, x_{n}$ such that it can be written as: $a(x - x_{1})*(x - x_{2})*...*(x-x_n)$, in which a is the leading coefficient.

Has zeros −19 and −i

If -i is a zero, its conjugate i is also a zero. So

$f(x) = a(x - (-19))(x - (-i))(x - i) = a(x+19)(x+i)(x-i) = a(x+19)(x^2 - i^2) = a(x + 19)(x^2 + 1)$

Output of 40 when x=1

This means that when $x = 1, f(x) = 40$. We use this to find the leading coefficient a. So

$f(x) = a(x + 19)(x^2 + 1)$

$40 = a(20)(2)$

$40a = 40$

$a = 1$

The polynomial is:

$f(x) = (x + 19)(x^2 + 1)$